A study was conducted to determine if an expectant mother’s cigarette smoking has an effect on the bone mineral content of her otherwise healthy baby A sample of 77 newborns whose mothers smoked have a mean bone mineral content of Xbar smokers = 0.099 gm/cm with a standard deviation of 0.026 gm/cm. A sample of 161 newborns whose mothers did not smoke have a mean bone mineral content of Xbar nonsmokers = 0.091 gm/cm with a standard deviation of 0.025 gm/cm. Assume that the variance of the underlying populations are equal.
Calculate the t value for the difference between the means.
What is the p value (2 tailed) for the above analysis (hint Excel tdist( t, degrees of freedom, number of tails)
Ans:
pooled standard deviation=sqrt(((77-1)*0.026^2+(161-1)*0.025^2)/(77+161-2))
=0.02533
standard error for difference=0.02533*sqrt((1/77)+(1/161))=0.0035
df=77+161-2=236
Test statistic:
t=(0.099-0.091)/0.0035
t=2.28
p-value(2 tailed)=tdist(2.28,236,2)=0.0235
A study was conducted to determine if an expectant mother’s cigarette smoking has an effect on...
A study was conducted to determine whether an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child. A sample of 77 newborns whose mothers smoked during pregnancy has mean bone mineral content x-bar1 = 0.098 g/cm and standard deviation s1 = 0.026 g/cm; a sample of 161 infants whose mothers did not smoke has mean x-bar2 = 0.095 g/cm and standard deviation s2 = 0.025 g/cm. Assume that the underlying population variances...